Prolog: cut & fail http://www.learnprolognow.org © Patrick Blackburn, Johan Bos & Kristina Striegnitz

The Cut Backtracking is a characteristic feature of Prolog But backtracking can lead to inefficiency: Prolog can waste time exploring possibilities that lead nowhere It would be nice to have some control The cut predicate !/0 offers a way to control backtracking

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X).

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- p(X).

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- p(X).

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- p(X).

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). X=1 ?- p(X). X=1

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),d(X),e(X). X=1 ?- p(X). X=1;

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),d(X),e(X). X=1 ?- p(X). X=1;

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),d(X),e(X). X=1 X=1 ?- c(1),d(1),e(1). ?- p(X). X=1;

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),d(X),e(X). X=1 X=1 ?- c(1),d(1),e(1). ?- d(1), e(1). ?- p(X). X=1;

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),d(X),e(X). X=1 X=1 ?- c(1),d(1),e(1). ?- d(1), e(1). ?- p(X). X=1; †

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),d(X),e(X). X=1 X=1 X=2 ?- c(1),d(1),e(1). ?- d(1), e(1). ?- p(X). X=1; †

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),d(X),e(X). X=1 X=1 X=2 ?- c(1),d(1),e(1). ?- c(2),d(2),e(2). ?- d(1), e(1). ?- p(X). X=1; †

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),d(X),e(X). X=1 X=1 X=2 ?- c(1),d(1),e(1). ?- c(2),d(2),e(2). ?- d(1), e(1). ?- d(2), e(2). ?- p(X). X=1; †

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),d(X),e(X). X=1 X=1 X=2 ?- c(1),d(1),e(1). ?- c(2),d(2),e(2). ?- d(1), e(1). ?- d(2), e(2). ?- p(X). X=1; † ?- e(2).

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),d(X),e(X). X=1 X=1 X=2 ?- c(1),d(1),e(1). ?- c(2),d(2),e(2). ?- d(1), e(1). ?- d(2), e(2). ?- p(X). X=1; X=2 † ?- e(2).

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),d(X),e(X). ?- f(X). X=1 X=1 X=2 ?- c(1),d(1),e(1). ?- c(2),d(2),e(2). ?- d(1), e(1). ?- d(2), e(2). ?- p(X). X=1; X=2; † ?- e(2).

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),d(X),e(X). ?- f(X). X=1 X=1 X=3 X=2 ?- c(1),d(1),e(1). ?- c(2),d(2),e(2). ?- d(1), e(1). ?- d(2), e(2). ?- p(X). X=1; X=2; X=3 † ?- e(2).

Example: cut-free code p(X):- a(X). p(X):- b(X), c(X), d(X), e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),d(X),e(X). ?- f(X). X=1 X=1 X=3 X=2 ?- c(1),d(1),e(1). ?- c(2),d(2),e(2). ?- d(1), e(1). ?- d(2), e(2). ?- p(X). X=1; X=2; X=3; no † ?- e(2).

Adding a cut Suppose we insert a cut in the second clause: If we now pose the same query we will get the following response: p(X):- b(X), c(X), !, d(X), e(X). ?- p(X). X=1; no

Example: cut p(X):- a(X). p(X):- b(X),c(X),!,d(X),e(X). p(X):- f(X). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X).

Example: cut p(X):- a(X). p(X):- b(X),c(X),!,d(X),e(X). ?- p(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- p(X).

Example: cut p(X):- a(X). p(X):- b(X),c(X),!,d(X),e(X). ?- p(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- p(X).

Example: cut p(X):- a(X). p(X):- b(X),c(X),!,d(X),e(X). ?- p(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). X=1 ?- p(X). X=1

Example: cut p(X):- a(X). p(X):- b(X),c(X),!,d(X),e(X). p(X):- f(X). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),!,d(X),e(X). X=1 ?- p(X). X=1;

Example: cut p(X):- a(X). p(X):- b(X),c(X),!,d(X),e(X). p(X):- f(X). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),!,d(X),e(X). X=1 ?- p(X). X=1;

Example: cut p(X):- a(X). p(X):- b(X),c(X),!,d(X),e(X). p(X):- f(X). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),!,d(X),e(X). X=1 X=1 ?- c(1),!,d(1),e(1). ?- p(X). X=1;

Example: cut p(X):- a(X). p(X):- b(X),c(X),!,d(X),e(X). p(X):- f(X). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),!,d(X),e(X). X=1 X=1 ?- c(1), !, d(1), e(1). ?- p(X). X=1; ?- !, d(1), e(1).

Example: cut X X p(X):- a(X). p(X):- b(X),c(X),!,d(X),e(X). p(X):- f(X). a(1). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). X ?- a(X). ?- b(X),c(X),!,d(X),e(X). X=1 X=1 X ?- c(1), !, d(1), e(1). ?- p(X). X=1; ?- !, d(1), e(1). ?- d(1), e(1).

Example: cut † p(X):- a(X). p(X):- b(X),c(X),!,d(X),e(X). p(X):- f(X). b(1). b(2). c(1). c(2). d(2). e(2). f(3). ?- p(X). ?- a(X). ?- b(X),c(X),!,d(X),e(X). X=1 X=1 ?- c(1), !, d(1), e(1). ?- p(X). X=1; no ?- !, d(1), e(1). ?- d(1), e(1). †

What the cut does The cut only commits us to choices made since the parent goal was unified with the the left-hand side of the clause containing the cut For example, in a rule of the form q:- p1, … , pn, !, r1, … , rn. when we reach the cut it commits us: to this particular clause of q to the choices made by p1, … , pn NOT to choices made by r1, … , rn

Using Cut Consider the following predicate max/3 that succeeds if the third argument is the maximum of the first two max(X,Y,Y):- X =< Y. max(X,Y,X):- X>Y.

Using Cut Consider the following predicate max/3 that succeeds if the third argument is the maximum of the first two max(X,Y,Y):- X =< Y. max(X,Y,X):- X>Y. ?- max(2,3,3). yes ?- max(7,3,7).

Using Cut Consider the following predicate max/3 that succeeds if the third argument is the maximum of the first two max(X,Y,Y):- X =< Y. max(X,Y,X):- X>Y. ?- max(2,3,2). no ?- max(2,3,5).

The max/3 predicate What is the problem? There is a potential inefficiency Suppose it is called with ?- max(3,4,Y). It will correctly unify Y with 4 But when asked for more solutions, it will try to satisfy the second clause. This is completely pointless! max(X,Y,Y):- X =< Y. max(X,Y,X):- X>Y.

max/3 with cut With the help of cut this is easy to fix Note how this works: If the X =< Y succeeds, the cut commits us to this choice, and the second clause of max/3 is not considered If the X =< Y fails, Prolog goes on to the second clause max(X,Y,Y):- X =< Y, !. max(X,Y,X):- X>Y.

Green Cuts Cuts that do not change the meaning of a predicate are called green cuts The cut in max/3 is an example of a green cut: the new code gives exactly the same answers as the old version, but it is more efficient

Red Cuts Cuts that change the meaning of a predicate are called red cuts The cut in the revised max/3 is an example of a red cut: If we take out the cut, we don’t get an equivalent program Programs containing red cuts Are not fully declarative Can be hard to read Can lead to subtle programming mistakes

Red Cuts ≠ if_then_else(P, Q, R) :- P, !, Q. if-then-else(P, Q, R) :- R. ≠ if_then_else(P, Q, R) :- P, Q. if-then-else(P, Q, R) :- R.

Another build-in predicate: fail/0 As the name suggests, this is a goal that will immediately fail when Prolog tries to proof it That may not sound too useful But remember: when Prolog fails, it tries to backtrack

Vincent and burgers The cut fail combination allows to code exceptions enjoys(vincent,X):- bigKahunaBurger(X), !, fail. enjoys(vincent,X):- burger(X). burger(X):- bigMac(X). burger(X):- bigKahunaBurger(X). burger(X):- whopper(X). bigMac(a). bigKahunaBurger(b). bigMac(c). whopper(d). The cut fail combination allows to code exceptions

Vincent and burgers The cut fail combination allows to code exceptions enjoys(vincent,X):- bigKahunaBurger(X), !, fail. enjoys(vincent,X):- burger(X). burger(X):- bigMac(X). burger(X):- bigKahunaBurger(X). burger(X):- whopper(X). bigMac(a). bigKahunaBurger(b). bigMac(c). whopper(d). ?- enjoys(vincent,a). yes The cut fail combination allows to code exceptions

Vincent and burgers The cut fail combination allows to code exceptions enjoys(vincent,X):- bigKahunaBurger(X), !, fail. enjoys(vincent,X):- burger(X). burger(X):- bigMac(X). burger(X):- bigKahunaBurger(X). burger(X):- whopper(X). bigMac(a). bigKahunaBurger(b). bigMac(c). whopper(d). ?- enjoys(vincent,b). no The cut fail combination allows to code exceptions

Vincent and burgers The cut fail combination allows to code exceptions enjoys(vincent,X):- bigKahunaBurger(X), !, fail. enjoys(vincent,X):- burger(X). burger(X):- bigMac(X). burger(X):- bigKahunaBurger(X). burger(X):- whopper(X). bigMac(a). bigKahunaBurger(b). bigMac(c). whopper(d). ?- enjoys(vincent,c). yes The cut fail combination allows to code exceptions

Vincent and burgers The cut fail combination allows to code exceptions enjoys(vincent,X):- bigKahunaBurger(X), !, fail. enjoys(vincent,X):- burger(X). burger(X):- bigMac(X). burger(X):- bigKahunaBurger(X). burger(X):- whopper(X). bigMac(a). bigKahunaBurger(b). bigMac(c). whopper(d). ?- enjoys(vincent,d). yes The cut fail combination allows to code exceptions

Negation as Failure The cut-fail combination seems to be offering us some form of negation It is called negation as failure, and defined as follows: neg(Goal):- Goal, !, fail. neg(Goal).

Vincent and burgers revisited enjoys(vincent,X):- burger(X), neg(bigKahunaBurger(X)). burger(X):- bigMac(X). burger(X):- bigKahunaBurger(X). burger(X):- whopper(X). bigMac(a). bigKahunaBurger(b). bigMac(c). whopper(d).

Vincent and burgers revisited enjoys(vincent,X):- burger(X), neg(bigKahunaBurger(X)). burger(X):- bigMac(X). burger(X):- bigKahunaBurger(X). burger(X):- whopper(X). bigMac(a). bigKahunaBurger(b). bigMac(c). whopper(d). ?- enjoys(vincent,X). X=a X=c X=d

Another build-in predicate: \+ Because negation as failure is so often used, there is no need to define it In standard Prolog the prefix operator \+ means negation as failure So we could define Vincent`s preferences as follows: ?- enjoys(vincent,X). X=a X=c X=d enjoys(vincent,X):- burger(X), \+ bigKahunaBurger(X).

Negation as failure and logic Negation as failure is not logical negation Changing the order of the goals in the vincent and burgers program gives a different behaviour: ?- enjoys(vincent,X). no enjoys(vincent,X):- \+ bigKahunaBurger(X), burger(X).

Summary of this lecture In this lecture we introduced cut & fail to control backtracking in Prolog.